El e c t ro nic J
o f
Pr
ob a bi l i t y
Electron. J. Probab.17(2012), no. 42, 1–17.
ISSN:1083-6489 DOI:10.1214/EJP.v17-2069
Distribution of the supremum location of stationary processes
∗Gennady Samorodnitsky
†Yi Shen
‡Abstract
The location of the unique supremum of a stationary process on an interval does not need to be uniformly distributed over that interval. We describe all possible distributions of the supremum location for a broad class of such stationary processes.
We show that, in the strongly mixing case, this distribution does tend to the uniform in a certain sense as the length of the interval increases to infinity.
Keywords:stationary process ; global supremum location ; bounded variation ; strong mixing.
AMS MSC 2010:60G10 ; 60G17.
Submitted to EJP on October 14, 2011, final version accepted on May 23, 2012.
1 Introduction
Let X = (X(t), t ∈ R)be a sample continuous stationary process. Even if, on an event of probability 1, the supremum of the process over a compact interval [0, T] is attained at a unique point, this point does not have to be uniformly distributed over that interval, as is known since [5]. However, its distribution still has to be absolutely continuous in the interior of the interval, and the density has to satisfy very specific general constraints, as was shown in a recent paper [7].
In this paper we give a complete description of the family of possible densities of the supremum location for a large class of sample continuous stationary processes. The necessary conditions on these densities follow by combining certain general results cited above, and for every function satisfying these necessary conditions we construct a stationary process of the required type for which this function is the density of the supremum location. This is done in Section 3, which is preceded by Section 2 in which we describe the class of stationary processes we are considering and quote the results from [7] we need in the present paper. Next, we show that for a large class of stationary processes, under a certain strong mixing assumption, the distribution of the supremum location does converge to the uniformity for very long intervals, and it does it in a strong sense. This is shown in Section 4.
∗Supported by grants ARO W911NF-07-1-0078, NSF DMS-1005903, NSA H98230-11-1-0154.
†Cornell University, USA. E-mail:[email protected]
‡Cornell University, USA. E-mail:[email protected]
2 Preliminaries
For most of this paper X = (X(t), t ∈ R) is a stationary process with continuous sample paths, defined on a probability space Ω,F, P
, but in Section 4 we will allow upper semi-continuous sample paths. In most of the paper (but not in Section 4) we will also impose two assumptions on the process, which we now state.
ForT >0we denote byX∗(T) = sup0≤t≤TX(t), the largest value of the process in the interval[0, T].
AssumptionUT: P
X(ti) =X∗(T), i= 1,2, for two differentt1, t2∈[0, T]
= 0.
Many processes satisfy AssumptionUT. In particular, a beautiful proof in [3] shows that any continuous Gaussian process, such thatX(s) 6=X(t)a.s. for any two points s6=t, satisfies this assumption.
The second assumption on a stationary process deals with the fluctuations of its sample paths.
AssumptionL:
K:= lim
ε↓0
P Xhas a local maximum in(0, ε)
ε <∞,
with the limit easily shown to exist. Under Assumption L the process X has sample paths of locally bounded variation; see Lemma 2.2 in [7].
For a compact interval[a, b], we will denote by τX,[a,b]= inf
t∈[a, b] : X(t) = max
a≤s≤bX(s)
the leftmost location of the supremum in the interval; it is a well defined random vari- able. If the supremum is unique, the adjective “leftmost” is, clearly, redundant. For a= 0, we will abbreviateτX,[0,b]toτX,b, and use the same abbreviation in similar situa- tions in the sequel.
We denote byFX,[a,b] the law ofτX,[a,b]; it is a probability measure on the interval [a, b]. It was proved in [7] that for anyT >0the probability measureFX,T is absolutely continuous in the interior of the interval[0, T], and density can be chosen to be right continuous and have left limits; we call this version of the densityfX,[a,b]. This version of the density satisfies a universal upper bound
fX,T(t)≤max 1
t, 1 T−t
, 0< t < T . (2.1)
We will also use the following result from the above reference.
Lemma 2.1. Let0≤∆< T. Then for every0≤δ≤∆,fX,T−∆(t)≥fX,T(t+δ)almost everywhere in(0, T −∆). Furthermore, for every suchδand everyε1, ε2≥0, such that ε1+ε2< T−∆,
Z T−∆−ε2 ε1
fX,T−∆(t)−fX,T(t+δ)
dt (2.2)
≤ Z ε1+δ
ε1
fX,T(t)dt+ Z T−ε2
T−∆−ε2+δ
fX,T(t)dt .
3 Processes satisfying Assumption L
In this section we prove our main theorem, giving a full description of possible càdlàg densitiesfX,Tfor continuous stationary processes satisfying Assumption UT and Assumption L.
For a functionf of a real argument whose domain contains an interval (t1, t2), its total variation over the interval is defined by
T V(t1,t2)(f) := sup
n−1
X
i=1
f(si+1)−f(si) ,
where the supremum is taken over all choices oft1< s1< . . . < sn< t2.
Theorem 3.1. LetX= (X(t), t∈R)be a stationary sample continuous process, sat- isfying Assumption UT and Assumption L. Then the restriction of the lawFX,T of the unique location of the supremum of the process in[0, T]to the interior(0, T)of the in- terval is absolutely continuous. The densityfX,T has a càdlàg version with the following properties:
(a) The density has a bounded variation on(0, T), hence the limits fX,T(0+) = lim
t→0fX,T(t)andfX,T(T−) = lim
t→TfX,T(t) exist and are finite. Furthermore,
T V(0,T)(fX,T)≤fX,T(0+) +fX,T(T−). (3.1) (b) The density is bounded away from zero. That is,
inf
0<t<TfX,T(t)>0. (3.2)
(c) EitherfX,T(t) = 1/T for all0< t < T, orRT
0 fX,T(t)dt <1.
Moreover, iffis a nonnegative càdlàg function satisfying (a)-(c) above, then there is a stationary sample continuous processX, satisfying Assumption UT and Assumption L, such thatfis the density in the interior(0, T)of the unique location of the supremum of the process in[0, T].
Proof. The existence of a càdlàg density with properties (a)-(c) in the statement of the theorem is an immediate consequence of the statements of Theorems 3.1 and 3.3 in [7]. We proceed to show the converse part of the theorem. If fX,T(t) = 1/T for all 0 < t < T, then a required example is provided by a single wave periodic stationary Gaussian process with period T, so we need only to consider the second possibility in property (c). We start with the case where the candidate density f is a piecewise constant function of a special form.
We call a finite collection(ui, vi), i= 1, . . . , mof nonempty open subintervals of(0, T) a proper collection of blocksif for anyi, j= 1, . . . , mthere are only 3 possibilities: either (ui, vi)⊆(uj, vj), or(uj, vj)⊆(ui, vi), or[ui, vi]∩[uj, vj] =∅. Ifui = 0, vi =T, we call (ui, vi)a base block. If ui = 0, vi < T, we call (ui, vi) a left block. If ui > 0, vi = T, we call(ui, vi)a right block. Ifui >0, vi < T, we call(ui, vi)a central block. We start with constructing a stationary process as required in the theorem when the candidate densityf satisfies requirements (a)-(c) of the theorem and has the form
f(t) = 1 HT
m
X
i=1
1[ui,vi)(t), 0< t < T (3.3)
for some proper collection of blocks, with the obvious convention at the endpoints 0 and T, for someH >1. Observe that for functions of the type (3.3), requirement (b) of the theorem is equivalent to requiring that there is at least one base block, and requirement (a) is equivalent to requiring that the number of the central blocks does not exceed the number of the base blocks. Finally, (the second case of) property (c) is equivalent to requiring that
d= 1
m HT −
m
X
i=1
(vi−ui)
!
>0. (3.4)
We will construct a stationary process by a uniform shift of a periodic deterministic function over its period. Now, however, the period will be equal to HT > T. We start, therefore, by defining a deterministic continuous function(x(t),0≤t≤HT)with x(0) = x(HT), which we then extend by periodicity to the entireR. LetB ≥1be the number of the base blocks in the collection. We partition the entire collection of blocks into B subcollection which we call components by assigning each base block to one component, assigning to each component at most one central block, and assigning the left and right blocks to components in an arbitrary way. Forj= 1, . . . , Bwe denote by
Lj=d the number of blocks in thejth component
(3.5) +the total length of the blocks in thejth component.
We setx(0) = 2. Using the blocks of the first component we will define the functionx on the interval(0, L1]in such a way thatx(L1) = 2. Next, using the blocks of the second component we will define the functionxon the interval(L1, L1+L2]in such a way that x(L1+L2) = 2, etc. Since
B
X
j=1
Lj=dm+
m
X
i=1
(vi−ui) =HT ,
this construction will terminate with a function x constructed on the entire interval [0, HT]withx(HT) = 2 =x(0), as desired.
We proceed, therefore, with defining the functionxon an interval of lengthLjusing the blocks of the jth component. For notational simplicity we will take j = 1 and definexon the interval[0, L1]using the blocks of the first component. The construction is slightly different depending on whether or not the component has a central block, whether or not it has any left blocks, and whether or not it has any right blocks. If the component hasl ≥1 left blocks, we will denote them by(0, vj),j = 1, . . . , l. If the component hasr ≥1 right blocks, we will denote them by(uj, T), j = 1, . . . , r. If the component has a central block, we will denote it by(u, v). We will construct the function xby defining it first on a finite number of special points and then filling in the gaps in a piecewise linear manner.
Suppose first that the component has a central block, some left blocks and some right blocks. In this case we proceed as follows.
Step 1 Recall thatx(0) = 2and set x jd+
j−1
X
i=1
vi
!
=x jd+
j
X
i=1
vi
!
= 2−2j−l, j= 1, . . . , l . Note that the last point obtained in this step isx ld+Pl
i=1vi
= 1. Step 2 Set
x (l+ 1)d+
l
X
i=1
vi
!
=x (l+ 1)d+
l
X
i=1
vi+v
!
=x (l+ 1)d+
l
X
i=1
vi+v+T−u
!
= 1 2. Step 3 Set
x (l+j+ 1)d+
l
X
i=1
vi+v+T −u+
j−1
X
i=1
(T−uj)
!
=x (l+j+ 1)d+
l
X
i=1
vi+v+T−u+
j
X
i=1
(T−uj)
!
= 2−2−(j−1), j= 1, . . . , r . Note that the last point obtained in this step is
x (l+r+ 1)d+
l
X
i=1
vi+v+T−u+
r
X
i=1
(T−uj)
!
= 2−2−(r−1).
Step 4 We add just one more point at distancedfrom the last point of the previous step by setting
x (l+r+ 2)d+
l
X
i=1
vi+v+T−u+
r
X
i=1
(T−uj)
!
= 2.
Note that this point coincides withL1as defined in (3.5).
If the component has no left blocks, then Step 1 above is skipped, and Step 2 be- comes the initial step with
x(d) =x(d+v) =x(d+v+T−u) =1 2.
If the component has no right blocks, then Step 3 above is skipped, and at Step 4 we add the distancedto the final point of Step 2, that is we set
x (l+ 2)d+
l
X
i=1
vi+v+T−u
!
= 2.
If the component has no central block, then Step 2 is skipped, but we do add the distanceT to the last point of Step 1. That is, the first point obtained at Step 3 becomes
x (l+ 1)d+
l
X
i=1
vi+T
!
= 1,
if there are any left blocks, with the obvious change ifl= 0. Finally, if there is neither central block, nor any right blocks, then both Step 2 and Step 3 are skipped, and Step 4 just addsd+T to the last point of Step 1, i.e. it becomes
x (l+ 1)d+
l
X
i=1
vi+T
!
= 2,
once again with the obvious change ifl= 0. It is easy to check that in any case Step 4 setsx(L1) = 2, withL1as defined in (3.5). In particular,L1> T.
Finally, we specify the piecewise linear rule by which we complete the construction of the functionxon the interval[0, L1]. The function has been defined on a finite set of points and we proceed from left to right, starting withx(0) = 2, to fill the gap between one point in the finite set and the adjacent point from the right, until we reachx(L1) = 2. By the construction, there are pairs of adjacent points in which the values ofxcoincide, and pairs of adjacent points in which the values ofxare different. In most cases only adjacent points at the distance d have equal values of x, but if, e.g. a central block is missing, then at a pair of adjacent points at a distance T, or d+T, the values of x coincide as well.
In any case, if the values of xat two adjacent points are different, we define the values ofxbetween these two points by linear interpolation. If the values ofxat two adjacent points, say,a and b witha < b, are equal to, say,y we define the function x between these two points by
x(t) = max y−(t−a)/d, y−(b−t)/d
provided the value at the midpoint,y−(b−a)/2d≥ −1. If this lower bound fails, we define the values ofxbetween the pointsa+dyandb−dyby
x(t) = max −τ(t−(a+dy)),−τ((b−dy)−t) ,
for an arbitraryτ > 0 such that bothτ ≤ 1/d and the value at the midpoint, −τ (b− a)/2−dy
≥ −1. The reason for this slightly cumbersome definition is the need to ensure thatxis nowhere constant, while keeping the lower bound ofxand its Lipschitz constant under control. We note, at this point, that, since in all casesb−a≤T+d, we can choose, for a fixedT, the value ofτ so thatτ ≥τd >0, where the constantτdstays bounded away from zero fordin a compact interval.
Now that we have defined a periodic function(x(t), t∈R)with periodHT, we define a stationary process X by X(t) = x(t−U), t ∈ R, where U is uniformly distributed between 0 andHT. The process is, clearly, sample continuous and satisfies Assumption L. We observe, further, that, if the supremum in the interval [0, T] is achieved in the interior of the interval, then it is achieved at a local maximum of the functionx. If the value at the local maximum is equal to 2, then it is due to an endpoint of a component, and, since the contribution of any component has length exceedingT, this supremum is unique. If the value at the local maximum is smaller than 2, then that local maximum is separated from the nearest local maximum with the same value ofxby at least the distance induced by Step 2, which T. Consequently, in this case the supremum over [0, T]is unique as well. Similarly, if the supremum is achieved at one of the endpoints of the interval, it has to be unique as well, on a set of probability 1. Therefore, the process Xsatisfies Assumption UT.
ExampleWe interrupt the exposition for a moment to demonstrate a simple special case of the construction of the processXto help the reader to visualize the procedure.
Consider a candidate density function f(t) =
2
HT ift∈(0, v1)∪[u, v),
1
HT ift∈[v1, u)∪[v, T)
for0 < v1< u < v < T, withH >1 +v1+v−uT . This corresponds to a proper collection of three blocks: a base block(0, T), a central block(u, v), and a left block(0, v1). Hence the total number of blocksm= 3, andd= 13(HT −T−v1−(v−u))>0. Since there is only one base block, we use one component, of the lengthL1=HT.
The construction of the deterministic function x(t) on [0, HT] is as follows. The starting point isx(0) = 2. Then step 1, dealing with the left block(0, v1), assigns value
2−21−1= 1to pointsdandd+v1. Step 2 continues to set
x(2d+v1) =x(2d+v1+v) =x(2d+v1+v+T−u) = 1 2.
Step 3 is skipped since there is no right block. Finally, the end point of this component, x(3d+v1+v+T−u) = 2is added in step 4. Since3d+v1+v+T−u=HT, this is the end of the cycle.
To demonstrate the the piecewise linear interpolation rule between these special points, we choose specific valuesT = 6, H = 2, v1 = 1, u = 3and v = 5. This implies d= 1. Firstly, between the pairs of points with thetcoordinates0andd= 1,d+v1= 2 and 2d+v1 = 3, 2d+v1+v+T −u= 11 and HT = 12we use linear interpolation.
Consider the segment between the points d = 1 and d+v1 = 2, at which x has the common value y = 1. The general rule checks the value of the interpolation at the midpoint of the segment, which is1−v2d1 = 1/2. It is greater than−1, so no modification is necessary. Same procedure applies to the segments between the points2d+v1 = 3 and2d+v1+v= 8, and the points2d+v1+v = 8and2d+v1+v+T−u= 11. Only on the interval(3,8)the interpolation procedure has to be modified. We setτ= 1/2(so that the value of the lowest point is exactly−1) and obtain
x(t) =
3.5−t if3≤t≤3.5 (3.5−t)/2 if3.5≤t≤5.5 (7.5−t)/2 if5.5≤t≤7.5 7.5−t if7.5≤t≤8
.
The figure below shows the densityf and the functionx(t)within one cycle.
0 1 2 3 4 5 6
1 / 1 2 1 / 6
f(t) t
0 2 4 6 8 1 0 1 2
- 1
012
x(t) t
( 0 , 2 )
( 1 , 1 ) ( 2 , 1 )
( 1 . 5 , 0 . 5 ) ( 3 , 0 . 5 )
( 3 . 5 , 0 )
( 5 . 5 , - 1 ) ( 7 . 5 , 0 )
( 8 , 0 . 5 )
( 9 . 5 , - 1 )
( 1 1 , 0 . 5 ) ( 1 2 , 2 )
Fig 1. The functionsfandxin the special case.
Finally, we extend the function x(t) periodically with periodHT to the whole real line. The processXis then defined byX(t) =x(t−U), whereU is uniformly distributed between0andHT.
We now return to the general case considered in the theorem. We first show that for the processXconstructed above, the densityfX,T coincides with the functionf given in (3.3), with which the construction was performed. According to the above analysis, we need to account for the contribution of each local maximum of the functionxover its period to the densityfX,T. The local maxima may appear in Step 1 of the construction, and then they are due to left blocks. They may apear in Step 3 of the construction, and then they are due to right blocks. They may appear Step 2 of the construction,
and then they are due to central blocks. Finally, the points wherex has value 2 are always local maxima. We will see that they are due to base blocks. We start with the latter local maxima. Clearly, each such local maximum is, by periodicity, equal to one of theB values,Pi
j=1Lj−HT, i= 1, . . . , B. Theith of these points becomes the global maximum ofXover[0, T]if and only if
U ∈
HT −
i
X
j=1
Lj,(H+ 1)T−
i
X
j=1
Lj
, and the global maximum is then located at the pointPi
j=1Lj−HT+U. Therefore, the contribution of each such local maximum to the density is1/HT at each0< t < T, and overall the points wherexhas value 2 contribute tofX,T
fbase(t) = B
HT, 0< t < T . (3.6)
Next, we consider the contribution tofX,Tof the local maxima due to left blocks. For simplicity of notation we consider only the left blocks in the first component. Then the local maximum due to thejth left block is at the pointjd+Pj
i=1vi. As before, we need to check over what interval of the values ofU this local maximum becomes the global maximum ofXover[0, T]. The relevant values ofU must be such that the time interval
jd+Pj−1
i=1vi, jd+Pj i=1vi
is shifted to cover the origin, and this corresponds to an interval of lengthvj of the values of U. The shifted local maximum itself will then be located within the interval(0, vj), which contributes1/HT at each0 < t < vj. Overall, the local maxima due to left blocks contribute tofX,T
fleft(t) = 1 HT
X
left blocks
1(0,vi)(t), 0< t < T . (3.7) Similarly, the local maxima due to right blocks contribute tofX,T
fright(t) = 1 HT
X
right blocks
1(ui,T)(t), 0< t < T . (3.8) Finally, we consider the central blocks. If the first component has a central block, then the local maximum due to the central block is at the point(l+ 1)d+Pl
i=1vi+v. Any value ofU that makes this local maximum the global maximum over [0, T] must be such that the time interval (l+ 1)d+Pl
i=1vi,(l+ 1)d+Pl
i=1vi+v
is shifted to cover the origin. Furthermore, that value ofU must also be such that the time interval
(l+1)d+Pl
i=1vi+v,(l+1)d+Pl
i=1vi+v+T−u
is shifted to cover the right endpointT. If we think of shifting the origin instead of shiftingx, the origin will have to be located in the interval (l+ 1)d+Pl
i=1vi,(l+ 1)d+Pl
i=1vi+v−u
. This corresponds to a set of values ofU of measurev−u, and the shifted local maximum will then be located within the interval(u, v), which contributes1/HT at eachu < t < vto the density. Overall, the local maxima due to central blocks contribute tofX,T
fcentral(t) = 1 HT
X
central blocks
1(u,v)(t), 0< t < T . (3.9) Since
fX,T(t) =fbase(t) +fleft(t) +fright(t) +fcentral(t), 0< t < T ,
we conclude by (3.6) - (3.9) that fX,T indeed coincides with the functionf given in (3.3). Therefore, we have proved the converse part of the theorem in the case when the candidate densityf is of the form (3.3).
We now prove the converse part of the theorem for a generalf with properties (a)-(c) in the statement of the theorem. Recall that we need only to treat the second possibility in property (c). In order to construct a stationary processXfor whichfX,T =f, we will approximate the candidate densityf by functions of the form (3.3). Since we will need to deal with convergence of a sequence of continuous stationary processes we have just constructed in the case when the candidate density is of the form (3.3), we record, at this point, several properties of the stationary periodic processX(t) =x(t−U),t∈R constructed above.
Property 1 The processXis uniformly bounded:−1≤X(t)≤2for allt∈R. Property 2 The processXis Lipschitz continuous, and its Lipschitz constant does not exceed3/2d.
Property 3 The process X is differentiable except at countably many points, at whichXhas left and right derivatives. On the setD0 ={t: X(t)>0}the derivatives satisfy
|X0(t)| ≥ 1 2Nd
(where the bound applies to both left and right derivatives iftis not a differentiability point). HereN is the bigger of the largest number of left blocks any component has, and the largest number of the right blocks any component has. Similarly, on the set D1={t: X(t)≤0}the derivatives satisfy
|X0(t)| ≥τd,
whereτd>0stays bounded away from zero fordin a compact interval.
Property 4 The distance between any two local maxima of X cannot be smaller thand. At its local maxima, Xtakes values in a finite set of at most N+ 3 elements.
Moreover, the absolute difference in the values of the processXin two local maxima in the interval(0, T)is at least2−N, whereN is as above.
All these properties follow from the corresponding properties of the function xby considering the possible configuration of the blocks in a component.
We will now construct a sequence of approximations to a candidate density f as above. Letn= 1,2, . . .. It follows from the general properties of càdlàg functions (see e.g. [1]) that there is a finite partition0 =t0 < t1 < . . . < tk =T of the interval[0, T] such that
|f(s)−f(t)| ≤ 1
nT for allti≤s, t < ti+1,i= 0, . . . , k−1. (3.10) We define a piecewise constant functionf˜non(0, T)by setting, for eachi= 1, . . . , k, the value off˜n forti−1≤t < tito be
f˜n(t) = 1 knT max
j= 0,1, . . .: f(s)≥ j
knT for allti−1≤s < ti
. By definition and (3.10) we see that
f(t)− 2
nT ≤f˜n(t)≤f(t), 0< t < T . (3.11) Next, we notice that for everyi= 1, . . . , k−1 there are pointssi∈(ti−1, ti)andsi+1 ∈ (ti, ti+1)such that
f(si)−f(si+1) ≥
f˜n(ti−)−f˜n(ti) − 1
knT .
Therefore,
T V(0,T)( ˜fn)≤T V(0,T)(f) + 1
nT . (3.12)
We now define
fn(t) = ˜fn(t) + 1
nT 0< t < T .
Clearly, the functionfn is càdlàg, has bounded variation on(0, T)and is bounded away from zero. By (3.12),fnalso satisfies (3.1) sincef does. Finally, sinceRT
0 fX,T(t)dt <1, we see by (3.11) that, for allnlarge enough, RT
0 fX,T(t)dt < 1as well. Therefore, for suchnthe functionfn has properties (a)-(c) in the statement of the theorem, and in the sequel we will only considern large as above. We finally notice thatfn takes finitely many different values, all of which are in the set{j/knT, j= 1,2, . . .}. Therefore,fncan be written in the form (3.3), withH=kn. Indeed, the blocks can be built by combining into a block all neighboring intervals where the value offn is the smallest, subtracting 1/knT from the value offn in the constructed block and iterating the procedure.
We have already proved that for any function of the type (3.3) there is a stationary process required in the statement of the theorem. Recall that a construction of this sta- tionary process depends on assignment of blocks in a proper collection to components, and we would like to make sure that no component has “too many” left or right blocks.
To achieve this, we need to distribute the left and right blocks as evenly as possible between the components. Two observations are useful here. First of all, it follows from the definition offnand (3.3) that
1
knnT(Ln+Bn) =fn(0+)≤f(0+) + 1
knnT ≤f(0+) + 1
fornlarge enough (we are writingkn instead of k to emphasize the dependence ofk on n), where Ln and Bn are the numbers of the the left and base blocks in the nth collection. On the other hand, similar considerations tell us that
1
knnTBn = inf
0<t<Tfn(t)≥ inf
0<t<Tf(t)− 2 nT ≥ 1
2 inf
0<t<Tf(t),
once again forn large enough, where we have used property (b) off. Therefore, for suchn,
Ln Bn
≤2 f(0+) + 1
inf0<t<Tf(t), (3.13)
and the right hand side is a finite quantity depending on f, but not on n. Performing a similar analysis for the right blocks, and recalling that we are distributing the left and right blocks as evenly as possible between the components, we see that there is a number∆f ∈(0,∞)such that for allnlarge enough, no component in thenth collection has more than∆f left blocks or∆f right blocks.
We will also need bounds on the important parameterd=dn appearing in the con- struction of a stationary process corresponding to functions of the type (3.3); these bounds do not depend on a particular way we assigns blocks to different components.
Recall that
dn= knnT mn
1− Z T
0
fn(t)dt
!
, (3.14)
wheremn =Bn+Ln+Rn+Cn(in the obvious notation) is the total number of blocks in thenth collection. Since
1
knnT Bn+ max(Ln, Rn, Cn)
= sup
0<t<T
fn(t), 1
knnTBn= inf
0<t<Tfn(t),
we see that
inf
0<t<Tfn(t)≤ 1
knnTmn≤3 sup
0<t<T
fn(t). (3.15)
We also know by the uniform convergence thatRT
0 fn →RT
0 f. Therefore, by (3.14) and (3.15) we obtain that, for allnlarge enough,
1−RT 0 f(t)dt
4 sup0<t<Tf(t) ≤dn≤ 2
inf0<t<Tf(t). (3.16) An immediate conclusion is the following fact. By construction, the distribution of Xn(0)is absolutely continuous; letgn denote the right continuous version of its density.
SinceXn is obtained by uniform shifting of a piecewise linear periodic function with periodHnT, the value of the densitygn(v)at each pointvtimes the length of the period does not exceed the total number of the linear pieces in a period divided by the smallest absolute slope of any linear piece. The former does not exceed2mn, and byProperty 3and the above, the latter cannot be smaller than
min 1
2∆fdn
, τdn
.
Since, by (3.16), dn is uniformly bounded from above, we conclude, for some finite positive constantc=c(f),gn(v)≤c(f)mn/Hn. Further, by the definition ofdn,
mndn=HnT P τXn,T ∈ {0, T}
≤HnT .
Once again, since by (3.16),dn is uniformly bounded from below, we conclude that gn(v)is uniformly bounded invandn. (3.17) Let Xn be the stationary process corresponding tofn constructed above. We view Xn as a random element of the space C(R) of continuous functions on R which we endow with the metric
ρ(x,y) =
∞
X
m=1
2−m sup
|t|≤m
|x(t)−y(t)|
.
Letµn be the law ofXnonC(R),n= 1,2, . . .(but large enough, as needed). ByProp- erty 1and Property 2of the processesXn and the lower bound in (3.16), these pro- cesses are uniformly bounded and equicontinuous. Therefore, by Theorem 7.3 in [1], for every fixedm= 1,2, . . .the restrictions of the measuresµnto the interval[−m, m]form a tight family of probability measures. Let n1j → ∞ be a sequence positive integers such that the restrictions ofµn1j to[−1,1]converge weakly to a probability measure ν1 on C([−1,1]). Inductively define for m = 2,3, . . . nmj → ∞ to be a subsequence of the sequence nm−1,j → ∞ such that the restrictions of µnmj to[−m, m] converge weakly to a probability measureνm on C([−m, m]). Then the “diagonal” sequence of measures µnjj, j = 1,2, . . .
is such that the restrictions of these measures to each interval[−m, m]converge weakly to νm onC([−m, m]). By the Kolmogorov existence theorem, there is a (cylindrical) probability measure ν on functions on R whose re- strictions to each interval [−m, m] coincide withνm (considered now as a cylindrical measure). Since each probability measure νm is supported by C([−m, m]), the mea- sureν itself is supported by functions inC(R). By construction, the measureν is shift invariant. If X is the canonical stochastic process defined on C(R), ν
, then X is a sample continuous stationary process. In the remainder of the proof we will show that Xsatisfies Assumption L and Assumption UT, and thatfX,T =f.
We start with proving that Assumption L holds forX. It is, clearly, enough to prove that, on a set of probability 1,
any two local maxima ofXare at leastθ:= 1−RT 0 f(t)dt
5 sup0<t<Tf(t) apart. (3.18) Suppose that (3.18) fails. Then there ismsucht that, on an event of positive probabil- ity, two local maxima ofXcloser thanθ exist in the time interval[−m, m]. Recall that a subsequence of the sequence of the (laws of) Xn converges weakly in the uniform topology onC([−m, m])to the (law of)X. For notational simplicity we will identify that subsequence with the entire sequence(Xn). By the Skorohod representation theorem (Theorem 6.7 in [1]), we may define the processes(Xn)on some probability space so thatXn→Xa.s. inC([−m, m]). Fixωfor which this convergence holds, and for which Xhas two local maxima closer thanθexist in the time interval[−m, m]. It straightfor- ward to check that the uniform convergence andProperty 3above imply that for alln large enough, the processesXnwill have two local maxima closer than5θ/4. This is, of course, impossible, due toProperty 4and (3.16). The resulting contradiction proves thatXsatisfies Assumption L.
Next, we prove that Assumption UT holds for X. Since the process X satisfies Assumption L, by Lemma 2.2 in [7], it has finitely many local maxima in the interval (0, T)(in fact, by (3.18), it cannot have more than dT /θelocal maxima). Clearly, the values ofXat the largest local maximum and the second largest local maximum (if any) are well defined random variables. We denote by(M1, M2)the largest and the second largest among X(0), X(T) and the values ofX at the largest local maximum and the second largest local maximum (if any). The fact that Assumption UT holds for X will follow once we prove that
P M1=M2
= 0. (3.19)
We proceed similarly to the argument in the proof of Assumption L. We may assume that Xn → X a.s. in C[0, T]. Fixω for which this convergence holds. The uniform convergence andProperty 3of the processes (Xn), together with the uniform upper bound ondnin (3.16), show that, for every local maximumtωofXin the interval(0, T) and anyδ >0, there isn(ω, δ)such that for alln > n(ω, δ), the processXn has a local maximum in the interval(tω−δ, tω+δ). This immediately implies that
M1−M2≥lim sup
n→∞
M1(n)−M2(n)
a.s., where the random vector(M1(n), M2(n))is defined for the processXn in the same way as the random vector(M1, M2)is defined for the processX,n= 1,2, . . .. In partic- ular, for anyε >0,
P M1−M2< ε
≤lim sup
n→∞
P M1(n)−M2(n)< ε
. (3.20)
As a first step, notice that, byProperty 4of the processes(Xn), for anyε <∆f, P
M1(n)−M2(n)< ε, (3.21)
bothM1(n)andM2(n)achieved at local maxima
= 0
for eachn. Next, since byProperty 4, at its local maxima the processXn can take at most∆f+ 3possible values, we conclude by (3.17) that for allε >0,
P
M1(n)−M2(n)< ε, one ofM1(n) (3.22)
andM2(n)is achieved at a local maximum, and one at an endpoint
≤cfε , for some cf ∈ (0,∞). Finally, we consider the case when both M1(n) and M2(n) are achieved at the endpoints of the interval[0, T]. In that case, it is impossible that Xn
has a local maximum in (0, T), since that would force time 0 to belong to one of the decreasing linear pieces of the process due to left blocks, and time T to belong one of the increasing linear pieces of the process due to right blocks. By construction, the distance between any two points belonging to such intervals is larger than T. That forcesXn(t),0 ≤t ≤ T to consist of at most two linear pieces. ByProperty 3 of the processXn, in order to achieve|Xn(0)−Xn(T)| ≤ε, each block of the proper collection generatingXncontributes at most an interval of lengthε/min(1/(2∆dn), τdn)to the set of possible shiftsU. Recall that there aremn blocks in the collection. By the uniform bounds (3.16) we conclude that for allε >0,
P
M1(n)−M2(n)< ε, (3.23)
M1(n)andM2(n)achieved at the endpoints
≤ε mn
HnT
1
min(1/(2∆dn), τdn) ≤ε 1
dnmin(1/(2∆dn), τdn)≤˜cfε , for some˜cf ∈(0,∞).
Combining (3.20), (3.21), (3.22) and (3.23) we see that for allε >0small enough, P M1−M2< ε
≤(cf+ ˜cf)ε .
Lettingε↓0we obtain (3.19), so that the processXsatisfies Assumption UT.
It is now a simple manner to finish the proof of the theorem. Assume, once again, thatXn → Xa.s. inC[0, T]. Fix ω for which this convergence holds, and bothXand each Xn have a unique supremum in the interval [0, T]. It follows from the uniform convergence thatτXn,T →τX,T asn→ ∞. Therefore, we also have thatτXn,T ⇒τX,T
(weakly). However, by construction,fn(t)→f(t)for every0< t < T. This implies that f is the density ofτX,T, and the proof of the theorem is complete.
Remark 3.2. Recall that most of the work in the construction of the process in Theo- rem 3.1 was done for the second case of condition (c). Under assumptionsUT and L this is the only alternative to the uniform distribution of the location of the supremum, and it guarantees that the probability that the supremum is located at an endpoint of the interval is positive. The separation of the local maxima property (3.18) of the con- structed process, for example, shows that that process satisfies assumptionL. Under assumptionsUT andL, the only scenario leading to the uniform distribution is to have the global suprema of the process appear periodically with a period ofT:
P X(τX,[T ,2T]) =X(τX,T), τX,[T ,2T]−τX,T =T
= 1;
see [7]. It is an open problem to describe all possible scenarios leading to the uniform distribution when Assumption L no longer holds, but the dichotomy of condition (c) in Theorem 3.1 no longer needs to hold in this case, as the example of the stationary Ornstein-Uhlenbeck process considered in [7] shows.
4 Long intervals
In spite of the broad range of possibilities for the distribution of the supremum location shown in the previous section, it turns out that, when the length of an interval
becomes large, and the process satisfies a certain strong mixing assumption, uniformity of the distribution of the supremum location becomes visible at certain scales. We make this statement precise in this section.
In this section we allow a stationary processXto have upper semi-continuous, not necessarily continuous, sample paths. Moreover, we will not generally impose either Assumption UT, or Assumption L. Without Assumption UT, the supremum may not be reached at a unique point, so we will work with the leftmost supremum location defined in Section 2.
Recall that a stationary stochastic process X = (X(t), t ∈ R) is called strongly mixing (orα-mixing) if
supn
P(A∩B)−P(A)P(B)
: A∈σ X(s), s≤0
, B∈σ X(s), s≥to
→0ast→ ∞;
see e.g. [6], p. 195. Sufficient conditions on the spectral density of a stationary Gaus- sian process that guarantee strong mixing were established in [4].
LetXbe an upper semi-continuous stationary process. We introduce a “tail version”
of the strong mixing assumption, defined as follows.
AssumptionTailSM: there is a functionϕ: (0,∞)→Rsuch that
t→∞lim P sup
0≤s≤t
X(s)≥ϕ(t)
= 1
and
supn
P(A∩B)−P(A)P(B)
: A∈σ X(s)1(X(s)≥ϕ(t)), s≤0
, B∈σ X(s)1(X(s)≥ϕ(t)), s≥to
→0ast→ ∞.
It is clear that if a process is strongly mixing, then it also satisfies Assumption TailSM. The point of the latter assumption is that we are only interested in mixing properties of the part of the process “responsible” for its large values. For example, the process
X(t) =
Y(t) ifY(t)>1
Z(t) ifY(t)≤1 , t∈R,
whereY is a strongly mixing process such thatP(Y(0) >1) >0, and Zan arbitrary stationary process such thatP(Z(0)<1) = 1, does not have to be strongly mixing, but it clearly satisfies Assumption TailSM withϕ≡1.
We will impose one more assumption on the stationary processes we consider in this section. It deals with the size of the largest atom the distribution of the supremum of the process may have.
AssumptionA:
lim
T→∞sup
x∈R
P sup
t∈[0,T]
X(t) =x
= 0.
In Theorem 4.1 below Assumption A could be replaced by requiring Assumption UT
for allT large enough. We have chosen Assumption A instead since for many impor- tant stationary stochastic processes the supremum distribution is known to be atomless anyway; see e.g. [8] for continuous Gaussian processes and [2] for certain stable pro- cesses. The following sufficient condition for Assumption A is also elementary: suppose that the processXis ergodic. If for somea∈R,P supt∈[0,1]X(t) =x
= 0for allx > a andP(X(0)> a)>0, then Assumption A is satisfied.
Theorem 4.1. LetX= (X(t), t∈R)be a stationary sample upper semi-continous pro- cess, satisfying Assumption TailSM and Assumption A. The densityfX,T of the supre- mum location satisfies
lim
T→∞ sup
ε≤t≤1−ε
T fX,T(tT)−1
= 0 (4.1)
for every0< ε <1/2. In particular, the law ofτX,T/T converges weakly to the uniform distribution on(0,1).
Proof. It is obvious that (4.1) implies weak convergence of the law of τX,T/T to the uniform distribution. We will, however, prove the weak convergence first, and then use it to derive (4.1).
We start with a useful claim that, while having nothing to do with any mixing by itself, will be useful for us in a subsequent application of Assumption TailSM. LetTn, dn ↑ ∞, dn/Tn→0asn→ ∞. We claim that for anyδ∈(0,1),
P
δTn−dn≤τX,Tn≤δTn+dn
= 0. (4.2)
To see this, simply note that by (2.1), the probability in (4.2) is bounded from above by 2dn sup
δTn−dn≤t≤δTn+dn
fX,Tn(t)≤2dnmax 1
δTn−dn
, 1
(1−δ)Tn−dn
→0 asn→ ∞.
The weak convergence stated in the theorem will follow once we prove that for any rational number r ∈ (0,1), we have P τX,T ≤ rT
→ r as T → ∞. Let r = m/k, m, k∈ N, m < k be such a rational number. ConsiderT large enough so thatT > k2, and partition the interval[0, T]into subintervals
Ci=
(T+√ T)i
k,(T+√ T)i+ 1
k −√
T
, i= 0,1, . . . , k−1,
Di=
(T+√ T)i
k−√
T ,(T+√ T)i
k
, i= 1, . . . , k−1, and observe that by (4.2),
P τX,T ∈
k−1
[
i=1
Di
→0asT → ∞.
Therefore,
P τX,T ≤rT
=P
0≤i≤m−1max Mi,T ≥ max
m≤i≤k−1Mi,T
+o(1) (4.3)
asT → ∞, whereMi,T = supt∈C
iX(t),i= 0,1, . . . .k−1. Letϕbe the function given in Assumption TailSM. Then
P
0≤i≤m−1max Mi,T ≥ max
m≤i≤k−1Mi,T
(4.4)
=P
0≤i≤m−1max Vi,T ≥ max
m≤i≤k−1Vi,T
+o(1), whereVi,T = supt∈C
iX(t)1 X(t)> ϕ(√ T)
,i= 0,1, . . . .k−1.
Denote byGT the distribution function of each one of the random variablesVi,T, and letWi,T =GT(Vi,T),i= 0,1, . . . , k−1. It is clear that
P
0≤i≤m−1max Vi,T ≥ max
m≤i≤k−1Vi,T
(4.5)
=P
0≤i≤m−1max Wi,T ≥ max
m≤i≤k−1Wi,T
.
Notice, further, that by Assumption TailSM, for every0< wi<1, i= 0,1, . . . , k−1, lim
T→∞
P
Wi,T ≤wi, i= 0,1, . . . , k−1
−
k−1
Y
i=0
P
Wi,T ≤wi
= 0. (4.6) Let
D(T) = sup
x∈R
P sup
t∈C0
X(t) =x
+P sup
t∈C0
X(t)≤ϕ(√ T)
. By Assumption A,D(T)→0asT→ ∞. Since for every0< w <1,
w−D(T)≤P
W0,T ≤w
≤w ,
we conclude by (4.6) that the law of the random vector W0,T, . . . , Wk−1,T
converges weakly, as T → ∞, to the law of a random vector (U0, . . . , Uk−1) with independent standard uniform components. Since this limiting law does not charge the boundary of the set{(w0, w1, . . . , wk−1) : max0≤i≤m−1wi≤maxm≤i≤k−1wi}, we conclude by (4.3), (4.4) and (4.5) that
P τX,T ≤rT
→P max
0≤i≤m−1Ui≥ max
m≤i≤k−1Ui
=m/k=r , and so we have established the weak convergence claim of the theorem.
We now prove the uniform convergence of the densities in (4.1). Suppose that the latter fails for some0 < ε < 1/2. There are two possibilities. Suppose first that there is θ > 0, a sequenceTn → ∞ and a sequence tn ∈ [ε,1−ε] such that for every n, TnfX,Tn(tnTn) ≥ 1 +θ. By compactness we may assume that tn → t∗ ∈ [ε,1−ε] as n → ∞. By Lemma 2.1 and the regularity properties of the density, for every n and every0< τ, δ <1such that
1−(1−τ)/tn
+< δ <min τ /tn,1
(4.7) we have
TnfX,(1−τ)Tn(tn(1−δ)Tn)≥TnfX,Tn(tnTn)≥1 +θ . Sincetn→t∗, there is a choice of0< τ <1such that
1 +θ > 1
1−τ (4.8)
and, moreover, the range in (4.7) is nonempty for allnlarge enough. Furthermore, we can find0< a < b <1such that
1−(1−τ)/tn
+ < a < b <min τ /tn,1 for allnlarge enough. Therefore, for suchn
(1 +θ)(b−a)≤ Z b
a
TnfX,(1−τ)Tn(tn(1−δ)Tn)dδ
= 1 tnP
τX,(1−τ)Tn∈ (1−b)tnTn, (1−a)tnTn
→ 1
1−τ(b−a)
asn → ∞ by the already established weak convergence. This contradicts the choice (4.8) ofτ.
The second way (4.1) can fail is that there is0 < θ < 1, a sequenceTn → ∞and a sequencetn ∈[ε,1−ε]such that for everyn,TnfX,Tn(tnTn)≤1−θ. We can show that this option is impossible as well by appealing, once again, to Lemma 2.1 and using an argument nearly identical to the one described above. Therefore, (4.1) holds, and the proof of the theorem is complete.
The following corollary is an immediate conclusion of Theorem 4.1. It shows the uniformity of the limiting conditional distribution of the location of the supremum given that it belongs to a suitable subinterval of[0, T].
Corollary 4.2. Let X = (X(t), t ∈ R) be a stationary sample upper semi-continous process, satisfying Assumption TailSM and Assumption A. Let0< aT ≤a0T < b0T ≤bT <
T be such that
lim inf
T→∞
aT
T >0, lim sup
T→∞
bT
T <1, lim
T→∞
b0T−a0T bT−aT
=θ.
Then
lim
T→∞P
τX,T ∈ a0T, b0T
τX,T ∈ aT, bT
=θ .
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Acknowledgments.We would like to thank the anonymous referee for the useful com- ments that helped us to improve the presentation of the material in the paper.
